Question

Please solve it........ ​

Answer 1

Answer:

here your answer...................

Answer 2

AnswEr :

$1.) \: \dfrac{ \tan( \theta) }{1 - \cot(\theta) } + \dfrac{ \cot(\theta) }{1 - \tan(\theta) } = 1 + \tan(\theta) + \cot(\theta)$

$\implies\dfrac{ \tan( \theta) }{1 - \cot(\theta) } + \dfrac{ \cot(\theta) }{1 - \tan(\theta) }$

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• changing every term into tan (θ)

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$\implies\dfrac{ \tan( \theta) }{1 - \frac{1}{\tan(\theta)} } + \dfrac{ \frac{1}{\tan(\theta)} }{1 - \tan(\theta) }$

$\implies\dfrac{ \tan( \theta) }{ \frac{ \tan(\theta) - 1}{\tan(\theta)} } + \dfrac{ \frac{1}{\tan(\theta)} }{1 - \tan(\theta) }$

$\implies\dfrac{ \tan^{2} ( \theta) }{ \tan(\theta) - 1} + \dfrac{1}{ \tan(\theta)( 1 - \tan(\theta)) }$

$\implies\dfrac{ \tan^{2} ( \theta) }{ \tan(\theta) - 1} - \dfrac{1}{ \tan(\theta)(\tan(\theta) - 1)}$

$\implies \dfrac{1}{\tan(\theta) - 1} \bigg( \tan^{2} ( \theta) - \dfrac{1}{ \tan(\theta)} \bigg)$

$\implies \dfrac{1}{\tan(\theta) - 1} \bigg( \dfrac{\tan^{3} ( \theta) - 1}{ \tan(\theta)} \bigg)$

$\implies \dfrac{1}{\tan(\theta) - 1} \bigg( \dfrac{\tan^{3} ( \theta) - {1}^{3} }{ \tan(\theta)} \bigg)$

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• (a³ - b³) = (a - b)(a² + ab + b²)

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$\implies \dfrac{1}{ \cancel{\tan(\theta) - 1}}\times \dfrac{ \cancel{(\tan(\theta) - 1)}(\tan^{2} ( \theta) + \tan( \theta) + 1) }{ \tan(\theta)}$

$\implies \dfrac{(\tan^{2} ( \theta) + \tan( \theta) + 1) }{ \tan(\theta)}$

$\implies \cancel\dfrac{\tan^{2}}{ \tan(\theta)} + \cancel\dfrac{\tan(\theta)}{\tan(\theta)} + \dfrac{1}{\tan(\theta)}$

$\implies \tan( \theta) + 1 + \cot(\theta)$

$\implies\boxed{1 + \tan( \theta)+\cot(\theta)}$

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$2.) \: \dfrac{1}{1 + \sin(A) } + \dfrac{1}{1 - \sin(A) } = 2 \sec^{2} (A)$

$\implies\dfrac{1}{1 + \sin(A) } + \dfrac{1}{1 - \sin(A) }$

$\implies\dfrac{1-\cancel{\sin(A)}+ 1 + \cancel{\sin(A)} }{(1 + \sin(A))(1 - \sin(A))}$

$\implies\dfrac{1 + 1}{1 - \sin^{2} (A)}$

$\implies\dfrac{2}{ \cos^{2} (A)}$

$\implies\boxed{2\sec^{2} (A)}$